Let $R$ be the region to the left of the line $x=\dfrac\pi 2$ and enclosed by that line, the $x$ -axis, and the curve $y=\text{sin}(x)$. $y$ $x$ ${y=\text{sin}(x)}$ $ R$ $ 0$ $\left(\dfrac\pi 2,1\right)$ $\dfrac\pi 2 $ A solid is generated by rotating $R$ about the line $x=\dfrac\pi 2$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi \int_0^{\frac\pi 2} \left[\text{sin}(y)-1\right]^2dy$ (Choice B) B $\pi \int_0^1 \left[\dfrac\pi 2-\text{sin}(y)\right]^2dy$ (Choice C) C $\pi \int_0^{\frac\pi 2} \left[\text{arcsin}(y)-1\right]^2dy$ (Choice D) D $\pi \int_0^1 \left[\dfrac\pi 2-\text{arcsin}(y)\right]^2dy$
Answer: Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=\text{sin}(x)}$ Notice the slices are horizontal, because we are rotating $R$ about a vertical axis. Each slice is a cylinder. Let the thickness of each slice be $dy$ and let the radius of the base, as a function of $y$, be $r(y)$. Then, the volume of each slice is $\pi [r(y)]^2\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [r(y)]^2\,dy$ This is called the disc method. What we now need is to figure out the expression of $r(y)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=\text{sin}(x)}$ $ 0$ $\left(\dfrac\pi 2,1\right)$ $\dfrac\pi 2 $ $r$ The radius is equal to the distance between the curve ${y=\text{sin}(x)}$ and the line ${x=\dfrac\pi 2}$. To find it, we need to solve $y=\text{sin}(x)$ for $x$ : ${x=\text{arcsin}(y)}$ So, for any $y$ -value, this is the equation for $r(y)$ : $\begin{aligned} r(y)}&={\dfrac\pi 2}-{\text{arcsin}(y)} \\\\ &=\dfrac\pi 2-\text{arcsin}(y)} \end{aligned}$ Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(y)}]^2 \\\\ &=\pi\left[\dfrac\pi 2-\text{arcsin}(y)}\right]^2 \end{aligned}$ The bottom endpoint of $R$ is at $y=0$ and the top endpoint is at $y=1$. So the interval of integration is $[0,1]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_0^1 \pi\left[\dfrac\pi 2-\text{arcsin}(y)\right]^2dy \\\\ &=\pi \int_0^1 \left[\dfrac\pi 2-\text{arcsin}(y)\right]^2dy \end{aligned}$